Question: Evaluate the double integral. $ \int_{0}^2 \int_{-y}^y x - 4y \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $-12$ (Choice B) B $\dfrac{-32}{3}$ (Choice C) C $\dfrac{-64}{3}$ (Choice D) D $\dfrac{-50}{3}$
First, we evaluate the inner integral. We can substitute in the $-y$ and $y$ at the end as if they were numerical bounds. $\begin{aligned} \int_{0}^2 \int_{-y}^y x - 4y \, dx \, dy &= \int_0^2 \left[ \dfrac{x^2}{2} - 4xy \right]_{-y}^y dy \\ \\ &= \int_0^2 \dfrac{y^2}{2} - 4y^2 - \left( \dfrac{y^2}{2} + 4y^2 \right) \, dy \\ \\ &= \int_0^2 -8y^2 \, dy \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^2 -8y^2 \, dy &= \left[ \dfrac{-8y^3}{3} \right]_0^2 \\ \\ &= \dfrac{-8(2)^3}{3} - \dfrac{-8(0)}{3} \\ \\ &= \dfrac{-64}{3} \end{aligned}$ The answer: $ \int_{0}^2 \int_{-y}^y x - 4y \, dx \, dy = \dfrac{-64}{3}$